What is the integral of (3x)/((1-x^2)^3) ?

The integral of (3x)/((1-x^2)^3) is 3/(4(1-x^2)^2)+C

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integral of (3x)/((1-x^2)^3)

\int \frac{3 x}{( 1 - x ^{2} ) ^{3}} d x

\frac{3}{4 ( 1 - x ^{2} ) ^{2}} + C

Solution steps

\int \frac{3 x}{( 1 - x ^{2} ) ^{3}} d x

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= 3 \cdot \int \frac{x}{( 1 - x ^{2} ) ^{3}} d x

Apply u-substitution

: \int - \frac{1}{2 u ^{3}} d u

= 3 \cdot \int - \frac{1}{2 u ^{3}} d u

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= 3 (- \frac{1}{2} \cdot \int \frac{1}{u ^{3}} d u)

Apply the Power Rule

: - \frac{1}{2 u ^{2}}

= 3 (- \frac{1}{2} (- \frac{1}{2 u ^{2}}))

Substitute back

u = 1 - x^{2}

= 3 (- \frac{1}{2} (- \frac{1}{2 ( 1 - x ^{2} ) ^{2}}))

Simplify

3 (- \frac{1}{2} (- \frac{1}{2 ( 1 - x ^{2} ) ^{2}})) : \frac{3}{4 ( 1 - x ^{2} ) ^{2}}

= \frac{3}{4 ( 1 - x ^{2} ) ^{2}}

Add a constant to the solution

= \frac{3}{4 ( 1 - x ^{2} ) ^{2}} + C

e^{y} d x + x e^{y} d y = 0

\frac{\partial}{\partial x} (cos^{2} (x))

(x^{2} (x^{2} + 9 x))^{6}

\frac{\partial}{\partial x} (ln (x + y^{2}))

\frac{\partial}{\partial x} (\frac{2 x y}{x + y})

What is the integral of (3x)/((1-x^2)^3) ?
The integral of (3x)/((1-x^2)^3) is 3/(4(1-x^2)^2)+C